Thread: complex numbers getting too complex!

Hi folks,

we were asked to show that $z^n - z^{-n} = 2i \sin nt$ ........... (1)

using de Moivre's theorem where n is a positive or negative integer and z is a complex number. This was ok.

then the question says, hence or otherwise prove that

$16 \sin^5 t = \sin 5t - 5 \sin 3t + 10 \sin t$ .........(2)

I had to resort to otherwise. I can see no way to get from (1) to (2)

Any help would be appreciated.

Thanks

$\displaystyle \text{sint} = \frac{z-z^{-1}}{2i}$

$\displaystyle \sin ^5t=\left( \frac{z-z^{-1}}{2i}\right)^5\text{=.}\text{..}$

expand using the binomial theorem

Many thanks Idea.

Originally Posted by s_ingram

Hi folks,

we were asked to show that $z^n - z^{-n} = 2i \sin nt$ ........... (1)

using de Moivre's theorem where n is a positive or negative integer and z is a complex number. This was ok.

I should mention that this is only true for z on the unit circle...

-Dan

Originally Posted by s_ingram

we were asked to show that $z^n - z^{-n} = 2i \sin nt$ ........... (1)
using de Moivre's theorem where n is a positive or negative integer and z is a complex number. This was ok.

I misread at first because (1) is not true.
Compare the results HERE with the results .

(1) is true for all $z$ for having the property $|z|=1$
In the example above, $|1+i|=\sqrt{2}~\&~\theta=\frac{\pi}{4}$

Where as $\left|\frac{1}{2}-\frac{\sqrt3}{2}i\right|=1~\&~\theta=\frac{-\pi}{3}$
Compare HERE to HERE (scroll down on the second one to compare)

Originally Posted by Idea

$\displaystyle \text{sint} = \frac{z-z^{-1}}{2i}$

$\displaystyle \sin ^5t=\left( \frac{z-z^{-1}}{2i}\right)^5\text{=.}\text{..}$

expand using the binomial theorem

Plato, thanks for your investigation, which I must admit is very puzzling. Certainly the formula provided gets me the right result, using Idea's idea.
But making the formula balance is another thing entirely. You have really confused me.
If I put z = 1 + i into the LHS of equation (1) I get 1/2 + 3/2i doing it manually and this matches your result for k=1. This number is a vector on the argand diagram magnitude $\dfrac{\sqrt{10}}{2}$ at angle $-\dfrac{\pi}{3}$ i.e. - 60 degrees. I would expect the right hand side to match, but $2 \sin \dfrac{\pi}{3}$ gives $- \sqrt{3} i$ which is, as you say wrong. I hope I am getting the wrong answer in the same way that you mean.

[QUOTE=s_ingram;934213]But making the formula balance is another thing entirely. You have really confused me.
If I put z = 1 + i into the LHS of equation (1) I get 1/2 + 3/2i doing it manually and this matches your result for k=1. This number is a vector on the argand diagram magnitude $\dfrac{\sqrt{10}}{2}$ at angle $-\dfrac{\pi}{3}$ i.e. - 60 degrees. I would expect the right hand side to match, but $2 \sin \dfrac{\pi}{3}$ gives $- \sqrt{3} i$ which is, as you say wrong. I hope I am getting the wrong answer in the same way that you mean.
My point was that $\displaystyle \text{sint} = \frac{z-z^{-1}}{2i}$] is true only if $|z|=1$
Therefore it is true for all points on the unit circle.

In general, $\forall z\ne 0$ it is true that $\large{z^{-1}=\frac{\overline{z}}{|z|^2}}$

ok. But even with this assumption, using z = 1 + i the left hand side doesn't match the right hand side! What is wrong with my calculation?

Originally Posted by s_ingram

ok. But even with this assumption, using z = 1 + i the left hand side doesn't match the right hand side! What is wrong with my calculation?

I really do not follow what side is which.
Because $|z=1+i|=\sqrt{2}\ne 0$ you cannot show $\dfrac{z-z^{-1}}{2i}=\sin\left( {\frac{\pi }{4}} \right)$

Can you say exactly what "using z = 1 + i the left hand side doesn't match the right hand side!" means?

Sorry if I have not been clear. I think we are in agreement actually. I am talking about the left hand side (LHS) and the right hand side of the equation $z^n - z^{-n} = i 2\sin t$

this should balance for any z, so if we take 1 + i then LHS = $\frac{1}{2} + i \frac{3}{2}$ for n = 1.

On the RHS we would have $i 2 \sin t$ and $t = \frac{\pi}{4}$ for z = 1 + i. i.e. $i \sqrt{2} $ so the LHS does not equal the RHS.

I was just trying to confirm that this is what you meant and that I hadn't made an error in my calculation.

Originally Posted by s_ingram

Sorry if I have not been clear. I think we are in agreement actually. I am talking about the left hand side (LHS) and the right hand side of the equation $z^n - z^{-n} = i 2\sin t$
this should balance for any z, so if we take 1 + i then LHS = $\frac{1}{2} + i \frac{3}{2}$ for n = 1.
On the RHS we would have $i 2 \sin t$ and $t = \frac{\pi}{4}$ for z = 1 + i. i.e. $i \sqrt{2} $ so the LHS does not equal the RHS.
I was just trying to confirm that this is what you meant and that I hadn't made an error in my calculation.

$ \begin{align*}z-z^{-1}&=z - \dfrac{1}{z}\\& = z - \dfrac{{\overline z }}{{{{\left| z \right|}^2}}} \\&= \dfrac{{|z{|^2}z - \overline z }}{{|z{|^2}}}\\&\mathop = \limits^? 2i\sin (Arg(z)) \end{align*}$

It should be clear that the last equality is only in case $|z|=1$

Therefore, it cannot balance for $z=1+i$

OK! I see what you mean.Thanks.